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allofthestars
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#1
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So if a feather takes longer to fall to the ground than a brick , its speed must be faster downwards and mass must be taken into consideration.

Then why do we use v= square root 2gh to calculate the impact speed, I understand this is the conservation of energy but surely for two objects such as the feather and brick their impact speeds would be different due to their masses ....yet masses cancel?
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Tubbz
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#2
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A feather falls slower than a brick because of wind resistance.
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xfcgdfg
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#3
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mgh=1/2mv^2
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Konanabanana
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#4
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mgh=1/2mv^2
Cancel Mass multiply both sides by 2
2gh=v^2
square root both sides
root 2gh=v


Look at your GPE KPE graph and you will see it is flipped should help you understand.
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black1blade
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Yeah that doesn't factor in air resistance.
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piketty6
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You use that if the question says "air resistance is negligible", because it means that the only force is weight hence it is only accelerating due to gravity.

As there is drag, you need to consider this force so you'd find the acceleration through F=ma and then use SUVAT.
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K-Man_PhysCheM
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(Original post by allofthestars)
So if a feather takes longer to fall to the ground than a brick , its speed must be faster downwards and mass must be taken into consideration.

Then why do we use v= square root 2gh to calculate the impact speed, I understand this is the conservation of energy but surely for two objects such as the feather and brick their impact speeds would be different due to their masses ....yet masses cancel?
The equation you are quoting is only valid in the absence of air resistance. Without air resistance, all objects in free fall accelerate at the same rate (9.8 m/s^2) , regardless of their mass, and so the impact speed of all objects in this idealised world without air resistance will be the same if they are dropped by the same distance.

Now in real life, air exists and applies a drag force on objects. A feather is light, so only needs a little upwards force to counteract the downwards weight force. Furthermore, a feather has a relatively large surface area for its weight, and as air drag increases with speed, it means that the feather only needs to be falling at quite a slow speed before the air drag force balances the weight force, so the feather reaches a slow terminal velocity (its fastest velocity on freefall).

However, as previously stated, the SUVAT equations and energy conservation does not take into account air drag (air drag causes energy to be lost as heat, so energy is still conserved, but instead of all being turned into kinetic energy, most is turned into heat energy in real life for a falling feather).

TLDR: that equation only works in an ideal world with no air resistance.
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Konanabanana
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(Original post by K-Man_PhysCheM)
The equation you are quoting is only valid in the absence of air resistance. Without air resistance, all objects in free fall accelerate at the same rate (9.8 m/s^2) , regardless of their mass, and so the impact speed of all objects in this idealised world without air resistance will be the same if they are dropped by the same distance.

Now in real life, air exists and applies a drag force on objects. A feather is light, so only needs a little upwards force to counteract the downwards weight force. Furthermore, a feather has a relatively large surface area for its weight, and as air drag increases with speed, it means that the feather only needs to be falling at quite a slow speed before the air drag force balances the weight force, so the feather reaches a slow terminal velocity (its fastest velocity on freefall).

However, as previously stated, the SUVAT equations and energy conservation does not take into account air drag (air drag causes energy to be lost as heat, so energy is still conserved, but instead of all being turned into kinetic energy, most is turned into heat energy in real life for a falling feather).

TLDR: that equation only works in an ideal world with no air resistance.

I think those who answered him know this. They don't make you consider other factors at A-Level.
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K-Man_PhysCheM
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(Original post by Konanabanana)
I think those who answered him know this. They don't make you consider other factors at A-Level.
When I started writing the answer, none of the answers published at the time stated that the SUVAT equations are only valid in the absence of resistive forces. I took a while to write my answer, and in that time a couple other people then wrote their own answers about this.

Notwithstanding, I am of the opinion that one should not confine themselves just to a specification (though of course, one should know their exam board's specification content thoroughly), and my extended answer was just written in case the OP was interested in the answer to their question, why does a feather NOT fall at the same speed as a brick in real life. If it's a little unnecessarily-detailed then so be it, but at least the OP may gain a better understanding of the subject. It's interesting, and moreover, you never know what sort of sneaky spec-stretching questions the A-level examiners may throw in there!
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Konanabanana
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(Original post by K-Man_PhysCheM)
When I started writing the answer, none of the answers published at the time stated that the SUVAT equations are only valid in the absence of resistive forces. I took a while to write my answer, and in that time a couple other people then wrote their own answers about this.

Notwithstanding, I am of the opinion that one should not confine themselves just to a specification (though of course, one should know their exam board's specification content thoroughly), and my extended answer was just written in case the OP was interested in the answer to their question, why does a feather NOT fall at the same speed as a brick in real life. If it's a little unnecessarily-detailed then so be it, but at least the OP may gain a better understanding of the subject. It's interesting, and moreover, you never know what sort of sneaky spec-stretching questions the A-level examiners may throw in there!
Fair enough I thought u were directing the message at me not the OP.
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allofthestars
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#11
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Thank you all for your amazing help.

I`m still rather confused though that , how come the velocity is limited to v= square root 2gh in a vacuum but the in earths atmosphere it can go far faster than say in a vacuum, like surely it would still be the same max velocity but just drag slows it down from reaching it as instantly as say on the moon ?
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K-Man_PhysCheM
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(Original post by allofthestars)
Thank you all for your amazing help.

I`m still rather confused though that , how come the velocity is limited to v= square root 2gh in a vacuum but the in earths atmosphere it can go far faster than say in a vacuum, like surely it would still be the same max velocity but just drag slows it down from reaching it as instantly as say on the moon ?
As you stated, in a vacuum v=\sqrt{2gh}.

That means that as the height fallen increases, the velocity increases too. If you fall 10m then v=\sqrt{2\times9.8\times10} = 14ms^{-1}, ie the object's speed after falling 10m is 14m/s. But if the object falls 100m, then v_{100} \approx 44.3ms^{-1}, which is a lot faster, and according to this equation, if an object falls an infinitely long distance then it will be travelling infinitely fast.

So your intuition is correct: objects fall much faster in a vacuum than in an atmosphere. This is because in a vacuum, no drag forces act on the object, so it never reaches terminal velocity and its speed keeps increasing indefinitely as it falls. On the other hand, in an atmosphere, drag forces cause objects to reach a terminal velocity.

Hope that helped
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Sam Keaney
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If you think of two identical umbrellas. If you drop one open and one closed at the same time the open one will hit the ground later and at a slower velocity than the closed one. Therefore the mass is irrelevant as the both have the same mass and the only variable is the surface area of the falling object and therefore, how much drag (air resistance) the object has.
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williamnguyen
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BOOOOOM mathematics rules. m cancels meaning it doesn't matter. DEAL WITH IT.
sqrt = square root
Just kidding. Let's talk about v= sqrt (2gh).
A wise man once said if something doesn't make sense, check your assumptions.
Well, you assumed no air resistance and hence KE = PE
And "v" used in the calculation is perhaps terminal velocity which is when weight = air resistance
--> mg = 1/2 k v^2 where (k is just a constant - a product of air density, etc look this up if you want)
thus v = sqrt (2mg/k) so clearly if you increase mass, you increase terminal velocity which means the impact will be greater due to a greater rate of change in
momemtum.

BOOOM So YE U HAVE IT. PHYSICS
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Muttley79
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(Original post by K-Man_PhysCheM)
When I started writing the answer, none of the answers published at the time stated that the SUVAT equations are only valid in the absence of resistive forces. I took a while to write my answer, and in that time a couple other people then wrote their own answers about this.

Notwithstanding, I am of the opinion that one should not confine themselves just to a specification (though of course, one should know their exam board's specification content thoroughly), and my extended answer was just written in case the OP was interested in the answer to their question, why does a feather NOT fall at the same speed as a brick in real life. If it's a little unnecessarily-detailed then so be it, but at least the OP may gain a better understanding of the subject. It's interesting, and moreover, you never know what sort of sneaky spec-stretching questions the A-level examiners may throw in there!
But in real-life a feather did fall for the same time as a hammer - this experiment was done on the moon.
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K-Man_PhysCheM
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(Original post by Muttley79)
But in real-life a feather did fall for the same time as a hammer - this experiment was done on the moon.
This was written over a year ago, so I'm not sure why you're bringing it up again.

But indeed, I am well aware of that experiment. I believe the thread of my argument is about how SUVAT equations are only valid in the absence of resistive forces. The OP wanted to know why in every-day experience, feathers and hammers don't fall for the same time. My argument went that in this situation, there are resistive forces. The constructive point is that if there are resistive forces, one cannot use the SUVAT equations since you no longer have constant acceleration. That answers the OPs question
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Muttley79
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(Original post by K-Man_PhysCheM)
This was written over a year ago, so I'm not sure why you're bringing it up again.

But indeed, I am well aware of that experiment. I believe the thread of my argument is about how SUVAT equations are only valid in the absence of resistive forces. The OP wanted to know why in every-day experience, feathers and hammers don't fall for the same time. My argument went that in this situation, there are resistive forces. The constructive point is that if there are resistive forces, one cannot use the SUVAT equations since you no longer have constant acceleration. That answers the OPs question
It's constant acceleration that is required ...
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K-Man_PhysCheM
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(Original post by Muttley79)
It's constant acceleration that is required ...
Ah, so we agree!

In the post you quoted, I wrote that SUVAT equations can NOT be used when acceleration is NOT constant. That is implicitly the same as what you have written!
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Muttley79
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(Original post by K-Man_PhysCheM)
Ah, so we agree!

In the post you quoted, I wrote that SUVAT equations can NOT be used when acceleration is NOT constant. That is implicitly the same as what you have written!
You implied no resistive forces though - there can be if they are constant.
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K-Man_PhysCheM
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#20
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(Original post by Muttley79)
You implied no resistive forces though - there can be if they are constant.
Ah, I see. Thank you for bringing that up. I was thinking about how in free-fall the drag force is proportional to speed or speed squared, at least according to the model used in A-level maths mechanics. So I was attempting to argue that in free-fall, resistive forces are not constant until terminal velocity is reached (hence SUVAT equations don't apply for the first part of the fall). But I didn't mention this specifically in any of my earlier posts.
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